Tag: shm as projection of circular motion

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Multiple choice physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

The particle executes SHM on a straight line. At two positions its velocity $u$ and $v$ while acceleration, $\alpha$ and $\beta$ respectively $[\beta > \alpha >0]$, the distance between the two positions will be:-

  1. $\frac{u^2+v^2}{\alpha+\beta}$

  2. $-\frac{u^2-v^2}{\alpha+\beta}$

  3. $\frac{u^2-v^2}{\alpha-\beta}$

  4. $\frac{u^2+v^2}{\beta-\alpha}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$u=w\sqrt{A^2-X _1^2}$
$V=w\sqrt{A^2-X _2^2}$
$\alpha=-w^2 x _1^2$
$\beta=-w^2 x _2^2$
$\frac{u^2-v^2}{u^2}=-(x _1^2-x _2^2)$
$=-(x _1+x _2)(x _1-x _2)$
$=\frac{(\alpha+\beta)}{w^2}(x _1-x _2)$
$x _1-x _2=\frac{u^2-v^2}{\alpha+\beta}$
Multiple choice physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

A stone is projected from the ground with a velocity of $14 \ ms^{-1}$ one second later it clean a wall $2 \ m$ high. The angle of projection is $(g = 10 \ ms^{-2})$ 

  1. $30 ^\circ$

  2. $45 ^\circ$

  3. $60 ^\circ$

  4. $15 ^\circ$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the equation of trajectory y = x*tan(theta) - (g*x^2)/(2*u^2*cos^2(theta)). With u=14, x = u*cos(theta)*t = 14*cos(theta)*1, and y=2, we solve for theta.

Multiple choice physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

The period of a particle it is $8s$. At $t=0$ it is at the mean position. The ratio of the distance covered by the particle in first second and second will be

  1. $\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } $

  2. $\cfrac { 1 }{ \sqrt { 2 } } $

  3. $\cfrac { 1 }{ \sqrt { 2 } -1 } $

  4. $\left[ \sqrt { 2 } -1 \right] $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x=A\sin\omega t$

$T= 8 sec$
$\dfrac{2\pi}{\omega}=8 sec$
$\omega=\dfrac{2\pi}{8}=\dfrac{\pi}{4}$
$x=A\sin \dfrac{\pi}{4} t$ ,At $t=1 sec$
$x _1=\dfrac{A}{\sqrt2}$ ,At $t=2sec$
$x _2=A$
Required ration $=\dfrac{x _1}{x _2-x _1}=\dfrac{1/\sqrt2}{1-\dfrac{1}{\sqrt2}}=\dfrac{1}{\sqrt2-1}$

Multiple choice physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

Sunlight of intensity 1.3 kW m–2 is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, kW m–2, at a distance 22 cm from the lens on the other side is _________. 

  1. $ 130 $

  2. $120 \ $

  3. $ 170$

  4. $160 $
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Multiple choice physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

Simple harmonic motion is the projection of uniform circular motion on the

  1. $x$- axis

  2. $y$- axis

  3. reference circle

  4. any diameter of reference circle.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Simple harmonic motion is the projection of uniform circular motion on any diameter of reference circle.